Suppose $|G|=p^nm$ with $1<m<p$. Show that $G$ is not simple.
I know it has to do with group actions. My idea is to consider a subgroup of order $p^n$, call it $H$. It has index $m$, so there are $m$ left cosets of $H$ in $G$, and $S_m$ acts on the set of left cosets. This gives a group homo $G\to S_m$. How do I proceed? I guess either $H$ should be normal or I should prove that the kernel of this map is a proper nontrivial subgroup, which is normal. But I don't know how to do either of those.